Re: NFB101 Crisis Averted
- From: Patrick Turner <info@xxxxxxxxxxxxxxxxxx>
- Date: Thu, 01 May 2008 15:40:07 GMT
Ian Thompson-Bell wrote:
Patrick Turner wrote:
Ian Thompson-Bell wrote:
Patrick Turner wrote:
Ian Thompson-Bell wrote:Actually, the input current is not zero; if it were the stage gain would
I finally cracked it - or rather found the text with an explanation IIndeed the the transconductance amp is one with extremely high input
could understand. Basically, with shunt applied feedback it is currents
that are summed/subtracted at the input. As the output is a voltage the
stage is a transconductance amplifier so the closed loop gain is in the
form of Vo/Ii which does turn out to be Ao/(1+ß.Ao) and ß is simply 1/Rf
where Rf is the feedback resistor
resistance.
So there is no input current to the amp, so a tube grid or fet gate
input
works well with shunt FB.
be infinite. There will always be an Rg from grid to cathode so an input
current i into the grid causes a Vgk of i.Rg
What I meant just to make things clearer is that the word 'amp' in my
above sentence
means the tube on its own, or any amp with extremely high Rin.
Yes there *is* current in the R1, but that's the current in the feedback
network only.
No, there is also the current from Vin.
You are disagreeing when you don't need to.
The current supplied from a voltage source signal flows in R1 and R2
equally. There is in effect, current flow between the amp input terminal
and the tube anode.
This one current determines grid voltage and the FB action we see.
Biasing a grid with shunt FB can be done from the signal input to
ground, and letting
whatever signal source provide the signal current into both the bias R
and FB network.
The shunt FB is more effective when biasing this way.
For ß to be equal to 1/Rf, then how come? where does the 1 come from?The transconductance amp does not have Ri present. ß is 1/Rf when
feeding current into the input (gri cct). If the open loop gain is high
then the closed loop gain is 1/ß or Rf so the output voltage is just i.Rf
I don't see you logic. Rf, or R2, could be any onld value so ß could be
anything.
Indeed. Remember, we are considering the amp *without* Ri, when it is a
transconductance amp. It's gain is Vo/Ii which must have the dimensions
of resistance (or impedance).
Are we considering an amp without R1?
I thought we were considering a tume or opamp ect always with R1 and R2
connected.
If you wish to share your problems then you MUST spell the set up we are
talking about.
Don't EVER assume I can read your mind. I'm a terrible mind reader.
If you said R1 was 1 ohm, i'd understand.
And I don't neglect the slight grid voltage at the input grid even if
the
ciruit is a high gain type.
?
What is it?no, some flows into Rg.
How come you don't express ß as Ri / ( Ri + Rf) , or in my way, R1 / (
R1 + R2 ) ?
The Ii or input current supplied by the voltage signal source flows
equally in R1 and R2.
But if Rg is from source to 0V, then input current is only flowing in
the FB network.
If you place Rg from g to 0V, everything becomed much more complex, and
I like simplicity
but the answer after using an equation must agree with simple analysis
using gain and ohm's
law.
And Kirchoff's law too.
But you have to know what the grid voltage is and Vo to sum these to getYes, it is i.Rg
the current in R2.
Or if you like, Ii = ( A + 1 ) / R2, where +A volts is at output and -1That I will do as part of NFB101.
volt is at the grid.
So the voltage across R1 = R1 x [ ( A + 1 ) / R2 ] = ( R1/R2 ) x ( A + 1
)
To get the input voltage you must add the R1 voltage to the grid voltage
which is 1, so you get
Vin = ( R1/R2 ) x ( A + 1 ) + 1.
closed loop A' = Vo / Vi. But Vo = +A volts, with -1V at the grid.
So A' = A / [ ( R1/R2 ) x ( A+1 ) + 1 ]
Consider a 1/2 6SN7, A = 16, R1 = 40k, R2 = 80k.
Say = +16 is at anode, then -1V is at grid, so voltage across 80k = 17V,
and across 40k is 8.5V, so input voltage MUST be -9.5V, ok?
A' = 16 / 9.5 .
Using the formula above without ß in it, just R1 and R2,
A' = 16 / [ 0.5 x (16 + 1) + 1 ] = 16 / 9.5
You should be able to prove whether your reasons and formulas work
with common types of triodes or pentodes,
and make sure other dumb folks will understand you.
Indeed, that voltage is i.RgThe input resistor effectively inputs a current equal to Vi/Ri fromThe input current is not Vin / R1.
which you can work out the voltage gain as Rf/Ri since the closed loop
transconductance is Rf. I'll include this in NFB101.
At the grid there is a smaller signal of the same phase as the input
signal.
So the current in R1 = ( Vin - Vg ) / R1.No. You can replace Vin and Ri with a Norton equivalent current
generator of Vin/Ri and parallel resistance Ri.
Yes but nobody would understand that so you need careful schematic
showing the basics.
Point taken. NFB101 *is* for relative beginners.
Assume the very worst about everyone out there trying to
design a simple triode line stage amp.
You do not need to mention Kirchoff, Norton.
Just use basics like I do and they'll all understand.
If you're going to write a paper on NFB, include the necessary 101
things,
not just 50, and never assume people will know the rest. They won't,
and they'll soon click off your pages once you have them bamboozled, and
they'll never return.
Include links to Norton and Kirchoff but being too heavy about these
will bamboozle.
All you need to know about building FB amp can be learnt by basic
current flow and ohm's law, and knowing how gain is developed in tubes.
Just my 2c opinion.
Patrick Turner.
.
Cheers
Ian
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